I am trying to prove the expression below
$$\iint\limits_R p(x) q(y)dxdy = \left( \int_a^bp(x) dx \right) \left( \int_c^dq(y) dy \right)$$
with $R = [a,b] \times [c,d]$
EDIT : $p(x)$ and $q(x)$ are continuous functions
I hope I made my self clear, english is not my first language. Thank you.
As $p$ and $q$ are continuous, so is the map $(x,y) \in R \mapsto F(x,y) = (p(x), q(y))$ (easy to prove). Now the map $(x,y) \in R \mapsto p(x) q(y)$ is just $B \circ F$, where $B: \Bbb R^2 \to \Bbb R$, $B(u,v) = uv$ (and $B$ is continuous, tell me if you need a proof). Hence, it is continuous. Applying Fubini's theorem,
$$\iint_R p(x)q(y) dx dy = \int_c^d \left[ \int_c^b p(x) q(y) dy \right] dx = \int_c^d p(x) \left[ \int_c^d q(y) dy \right] dx $$
the expression under brackets is independent of $x$, so it gets outside the whole integral and we get:
$$ \left( \int_a^b p(x) dx \right) \left( \int_c^d q(y) dy\right)$$